Angular Momentum in a pulley-block system

1. The problem statement, all variables and given/known data
A counterweight of mass m 5 4.00 kg is attached to
a light cord that is wound around a pulley.
The pulley is a thin hoop of radius R =
8.00 cm and mass M = 2.00 kg. The spokes have negligible
mass. When the counterweight has a speed v, the pulley
has an angular speed v = v/R. Determine the magnitude
of the total angular momentum of the system
about the axle of the pulley

2. Relevant equations

3. The attempt at a solution
I know the answer is L = rmv = (0.08)(2 + 4)v = 0.48v
but i don't understand why you have to add the momentum of the block using the equation of angular momentum I mean, it is not moving in a circle, it is moving in a straight line downward.

My original calculation was L = (0..08)(2)(v) + (4)(v)

Is this because when the block "transfers" its momentum to the pulley, its momentum will "act on" the surface of pulley at 0.08 m which is why I have to calculate it like (4)(v)(0.08)?

The "why" here is so that angular momentum is conserved. If you total up the angular momentum of all objects in a system, then let them bounce off each other, then total up the angular momentum again, you get the same number. That works only if you use the angular momentum formula.

Or to look at it another way, consider torque. If you compare the change in angular momentum that torque produces, it looks just like the change in linear momentum that a linear force produces.

## F = \frac {d (mv)}{dt}##

##T = \frac{d (I \omega)}{dt}##

And Newton's laws for linear motion, momentum, and force have direct parallels with angular motion, momentum, and torque. So all the rules and experience you have gained with linear momentum have direct parallels with angular momentum as well. Of course, there are a few places where there are sharp corners. Angular momentum is a pseudo vector, for example. And the "centre of angular momentum frame", if you were to use it, would probably be a rotating frame. And rotating frames have some surprising properties.

But essentially, you need this form of angular momentum to get the conservation of angular momentum.